The given system of equation is $\frac{2}{x} + \frac{3}{y} + \frac{10}{z}$ = 4 , $\frac{4}{x} - \frac{6}{y} + \frac{5}{z}$ = 1 , $\frac{6}{x} + \frac{9}{y} - \frac{20}{z}$ = 2 The given system of equation can be written as $\begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} \begin{bmatrix} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{bmatrix}$ = $\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$ or AX B,Where A = $\begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$ , X = $\begin{bmatrix} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{bmatrix}$ and B = $\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$ Now, |A| = $\begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) = 1200 ≠ 0 Hence, the unique solution of the system of equation is given by X = $A^{-1}B$ Now, the cofactors of A are computed as : $C_{11}$ = $(-1)^2$ (120 - 45) = 75, $C_{12}$ = $(-1)^3$ (- 80 - 30) = 110, $C_{13}$ = $(-1)^4$ (36 + 36) = 72 $C_{21}$ = $(-1)^3$ (- 60 - 90) = 150, $C_{22}$ = $(-1)^4$ (- 40 - 60) = - 100, $C_{23}$ = $(-1)^5$ (18 - 18) = 0 $C_{31}$ = $(-1)^4$ (15 + 60) = 75, $C_{12}$ = $(-1)^5$ (10 - 40) = 30, $C_{33}$ = $(-1)^6$ (- 12 - 12) = - 24 ∴ Adj A = $\begin{bmatrix} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{bmatrix}^T$ = $\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$ ⇒ $S^{-1}$ = $\frac{\operatorname{Adj}A}{|A|}$ = $\frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$ X = $A^{-1}B$ = = $\frac{1}{1200} \begin{bmatrix} 400+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix}$ = $\frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix}$ X = $\begin{bmatrix} \frac{600}{1200} \\ \frac{400}{1200} \\ \frac{240}{1200} \end{bmatrix}$ = $\begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix}$ ⇒ $\begin{bmatrix} \frac{1}{x} \\ \frac{1}{y} \\ \frac{1}{z} \end{bmatrix}$ = $\begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix}$ ⇒ $\frac{1}{x}$ = $\frac{1}{2}, \frac{1}{y}$ = $\frac{1}{3}$ and $\frac{1}{z}$ = $\frac{1}{5}$ ⇒ x = 2 , y = 3 and z = 5 Thus, solution of given system of equation is given by x = 2, y = 3 and z = 5. OR The given matrix is A = $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$ We have $AA^{-1}$ = I Thus, A = IA Or, $\begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ A Applying $R_2$ → $R_2 + 3R_1$ and $R_3$ → $R_3 - 2R_1$ $\begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix}$ A Now,applying $R_2$ → $\frac{1}{9} R_2$ $\begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & -\frac{7}{9} \\ 0 & -5 & 4 \end{bmatrix}$ = $\begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -2 & 0 & 1 \end{bmatrix}$ A Applying $R_1$ → $R_1 - 3R_2$ and $R_3$ → $R_3 + 5R_2$ $\begin{bmatrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{7}{9} \\ 0 & 0 & \frac{1}{9} \end{bmatrix}$ = $\begin{bmatrix} 0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -\frac{1}{3} & \frac{5}{9} & 1 \end{bmatrix}$ A Applying $R_3$ → $9R_3$ $\begin{bmatrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{7}{9} \\ 0 & 0 & 1 \end{bmatrix}$ = $\begin{bmatrix} 0 & -\frac{1}{3} & 0 \\ \frac{1}{3} & \frac{1}{9} & 0 \\ -3 & 5 & 9 \end{bmatrix}$ A Applying $R_1$ → $R_1 - \frac{1}{3} R_3$ and $R_2$ → $R_2 + \frac{7}{9} R_3$ $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ = $\begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix}$ A ⇒ I = $\begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix}$ A ∴ $A^{-1}$ = $\begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix}$ Hence, inverse of the matrix A is $\begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix}$